Find $\dfrac{d}{dx}\left[\sin(x)x^2\right]$. Choose 1 answer: Choose 1 answer: (Choice A) A $x^2\cos(x)+2x\sin(x)$ (Choice B) B $2x\cos(x)$ (Choice C) C $2x\cos(x)-x^2\sin(x)$ (Choice D) D $2x\sin(x)$
Answer: $\sin(x)x^2$ is the product of two, more basic, expressions: $\sin(x)$ and $x^2$. Therefore, the derivative of the expression can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[\sin(x)x^2\right] \\\\ &=\dfrac{d}{dx}\left[\sin(x)\right]x^2+\sin(x)\dfrac{d}{dx}[x^2]&&\gray{\text{The product rule}} \\\\ &=\cos(x)x^2+\sin(x)2x&&\gray{\text{Differentiate }\sin(x)\text{ and }x^2} \\\\ &=x^2\cos(x)+2x\sin(x)&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left[\sin(x)x^2\right]=x^2\cos(x)+2x\sin(x)$